Chapter 10: Logarithms — Part 3: Logarithmic Equations and Inequalities

CHAPTER 10: LOGARITHMS

PART 3: LOGARITHMIC EQUATIONS, INEQUALITIES AND FURTHER PROBLEM SOLVING

Learning objectives

By the end of this part, you should be able to:

1. Determine the values of a variable for which a logarithmic expression is defined.
2. Solve equations involving one or more logarithms.
3. Solve logarithmic equations that reduce to quadratic equations.
4. Solve equations involving logarithms with different bases.
5. Solve simple logarithmic inequalities.
6. Check solutions in the original equation and reject invalid values.

1. Conditions for logarithmic expressions

Before solving logarithmic equations, we must remember that a logarithm of a real number is defined only when its argument is positive. For a logarithm

log_a x

the conditions are:

• a > 0
• a ≠ 1
• x > 0

The expression inside the logarithm is called the argument. For example:

• log₂(x − 3) requires x − 3 > 0, hence x > 3.
• ln(2x + 1) requires 2x + 1 > 0, hence x > −½.
• log₅(x² − 4) requires x² − 4 > 0, hence x < −2 or x > 2.

Important examination rule: Always establish the domain before solving a logarithmic equation. An algebraic answer is not automatically a valid logarithmic solution.

Worked Example 1: Finding the domain

Find the values of x for which the expression is defined:

ln(x − 2) + ln(3x + 1)

Show complete solution

Both logarithmic arguments must be positive.

x − 2 > 0

Therefore, x > 2.

3x + 1 > 0

Therefore, x > −⅓. Both conditions must hold simultaneously. Their intersection is:

x > 2

Answer: The domain is (2, ∞).

Worked Example 2: A quadratic argument

Find the domain of:

log₃(x² − 5x + 6)

Show complete solution

The argument must be strictly positive:

x² − 5x + 6 > 0

Factorising:

(x − 2)(x − 3) > 0

The critical values are 2 and 3. The product is positive outside these two values.

x < 2   or   x > 3

Answer: (−∞, 2) ∪ (3, ∞).

2. Solving logarithmic equations

A logarithmic equation is an equation in which the unknown appears in the argument or base of a logarithm. The main strategy is:

1. State the restrictions on the logarithmic arguments.
2. Use the laws of logarithms to combine or simplify expressions.
3. Remove the logarithms when the bases are equal.
4. Solve the resulting algebraic equation.
5. Check every candidate in the original equation.

The following principle is especially useful:

If log_a M = log_a N, where both arguments are positive and a > 0, a ≠ 1, then:

M = N

Worked Example 3: Combining two logarithms

Solve:

log₂(x − 1) + log₂(x + 2) = 3

Show complete solution

Step 1: State the domain.

x − 1 > 0   and   x + 2 > 0

Hence x > 1. Step 2: Apply the product law.

log₂[(x − 1)(x + 2)] = 3

Step 3: Convert to index form.

(x − 1)(x + 2) = 2³

x² + x − 2 = 8

x² + x − 10 = 0

Step 4: Solve.

x = (−1 ± √41)/2

Step 5: Check the domain. Since x > 1, only the positive root is valid.

x = (−1 + √41)/2

Answer: x = (√41 − 1)/2.

Worked Example 4: Difference of logarithms

Solve:

log₃(2x + 1) − log₃(x − 1) = 1

Show complete solution

Step 1: Domain.

2x + 1 > 0,   x − 1 > 0

Hence x > 1. Step 2: Apply the quotient law.

log₃[(2x + 1)/(x − 1)] = 1

Step 3: Convert to index form.

(2x + 1)/(x − 1) = 3

2x + 1 = 3x − 3

x = 4

Since 4 is greater than 1, it satisfies the domain. Answer: x = 4. Check: The original left-hand side is log₃9 − log₃3 = 2 − 1 = 1.

3. Equations involving a quadratic in a logarithm

Some equations contain a logarithm raised to a power. These can often be solved by using substitution. For example, if an equation contains:

(log_ax)² + p(log_ax) + q = 0

we may put:

u = log_ax

The equation then becomes an ordinary quadratic in u.

Worked Example 5: Quadratic in a logarithm

Solve:

(log₂x)² − 3log₂x + 2 = 0

Show complete solution

Let u = log₂x.

u² − 3u + 2 = 0

Factorising:

(u − 1)(u − 2) = 0

Therefore:

u = 1   or   u = 2

Replacing u:

log₂x = 1   or   log₂x = 2

Converting to index form:

x = 2¹   or   x = 2²

Answer: x = 2 or x = 4. Both values are positive and satisfy the original equation.

Worked Example 6: A quadratic with a different base

Solve:

2(log₃x)² + log₃x − 3 = 0

Show complete solution

Put u = log₃x.

2u² + u − 3 = 0

Factorising:

(2u + 3)(u − 1) = 0

Hence:

u = −3/2   or   u = 1

Returning to the logarithm:

log₃x = −3/2   or   log₃x = 1

Therefore:

x = 3^−3/2   or   x = 3

Answer:

x = 1/(3√3)   or   x = 3

Both solutions are positive, as required.

4. Equations involving different logarithmic bases

When logarithms have different bases, the change-of-base law can be used to express them in a common base.

log_ax = (log_bx)/(log_ba)

This allows us to rewrite logarithms in terms of a common base.

Worked Example 7: Using change of base

Solve:

log₂x = log₄16

Show complete solution

First evaluate the right-hand side:

log₄16 = 2

Hence:

log₂x = 2

Converting to index form:

x = 2² = 4

Answer: x = 4.

Worked Example 8: An equation with two different bases

Solve:

log₂x + log₄x = 6

Show complete solution

Let u = log₂x. By change of base:

log₄x = log₂x / log₂4 = u/2

Therefore:

u + u/2 = 6

3u/2 = 6

u = 4

Thus:

log₂x = 4

x = 2⁴ = 16

Answer: x = 16.

5. Solving logarithmic inequalities

A logarithmic inequality compares two logarithmic expressions using signs such as:

<,   >,   ≤,   ≥

The important point is that the logarithmic function behaves differently depending on its base. Case 1: Base greater than 1 If a > 1, the function log_ax is increasing. Therefore:

log_aM > log_aN

implies:

M > N

provided both arguments are positive. Case 2: Base between 0 and 1 If 0 < a < 1, the function is decreasing. Therefore the inequality reverses:

log_aM > log_aN

implies:

M < N

again, provided both arguments are positive.

Remember: For bases greater than 1, the inequality direction stays the same. For bases between 0 and 1, it reverses when comparing the arguments.

Worked Example 9: Inequality with base greater than 1

Solve:

log₂(x + 1) > 2

Show complete solution

Domain:

x + 1 > 0   ⇒   x > −1

Since the base 2 is greater than 1, the inequality direction remains unchanged.

x + 1 > 2²

x + 1 > 4

x > 3

This satisfies the domain. Answer: x > 3.

Worked Example 10: Inequality with a base between 0 and 1

Solve:

log_1/2(x − 1) ≥ 2

Show complete solution

Domain:

x − 1 > 0   ⇒   x > 1

Since 0 < 1/2 < 1, the logarithmic function is decreasing. The inequality reverses when converting to the argument:

x − 1 ≤ (1/2)²

x − 1 ≤ 1/4

x ≤ 5/4

Combining with the domain:

1 < x ≤ 5/4

Answer: (1, 5/4].

Worked Example 11: An inequality involving ln

Solve:

ln(2x − 1) ≤ ln(x + 2)

Show complete solution

Domain conditions:

2x − 1 > 0   ⇒   x > 1/2

x + 2 > 0   ⇒   x > −2

Hence the combined domain is x > 1/2. Since the natural logarithm is increasing:

2x − 1 ≤ x + 2

x ≤ 3

Combining with the domain:

1/2 < x ≤ 3

Answer: (1/2, 3].

6. Further problem solving and common errors

The following points are important when answering examination questions:

• Do not take the logarithm of zero or a negative number in real-number work.
• Do not forget that every logarithmic argument must be positive.
• Do not combine logarithms with different bases using the product or quotient law until they have been expressed in a common base.
• When a quadratic is formed, check both roots against the original domain.
• For inequalities, establish whether the logarithmic base is greater than 1 or between 0 and 1.
• Use strict inequalities for logarithmic arguments: an argument cannot equal zero.

Worked Example 12: A complete equation with an invalid root

Solve:

ln(x − 1) + ln(x + 2) = ln 4

Show complete solution

Domain:

x − 1 > 0,   x + 2 > 0

Therefore x > 1. Apply the product law:

ln[(x − 1)(x + 2)] = ln 4

Equate the arguments:

(x − 1)(x + 2) = 4

x² + x − 2 = 4

x² + x − 6 = 0

Factorising:

(x + 3)(x − 2) = 0

Hence x = −3 or x = 2. However, the domain requires x > 1. We reject −3. Answer: x = 2. Check: ln 1 + ln 4 = 0 + ln 4 = ln 4.

7. Student exercises

Exercise A: Domains of logarithmic expressions

Find the values of x for which each expression is defined.

1. log₂(x − 5)
2. ln(4x + 8)
3. log₅(x² − 9)
4. ln[(x − 1)(x + 3)]

Exercise B: Logarithmic equations

Solve the following equations, giving all valid real solutions.

1. log₂x + log₂(x − 2) = 3
2. log₃(x + 1) − log₃(x − 1) = 1
3. (log₂x)² − 5log₂x + 6 = 0
4. 2(log₅x)² − 3log₅x − 2 = 0
5. log₂x + log₄x = 6
6. ln(x − 2) + ln(x + 1) = ln 4

Exercise C: Logarithmic inequalities

Solve each inequality and express your answer using inequality or interval notation.

1. log₃(x + 2) > 2
2. log_1/3(x − 1) > 1
3. ln(3x + 1) ≥ ln(x + 5)
4. log₂(x − 1) ≤ 3

Exercise D: Further problem solving

1. Solve log₂(x² − 2x) = 3, stating the domain first.
2. Solve log₃x + log₃(x − 6) = 2.
3. Solve [log₄x]² − 3log₄x + 2 = 0.
4. Solve ln(x + 3) − ln(x − 1) = ln 2.

8. Answers to exercises

Attempt the questions before opening the answers. For questions requiring full working, use the methods demonstrated in the worked examples.

Exercise A: Answers

Reveal answers

1. x > 5
2. x > −2
3. x < −3 or x > 3
4. x < −3 or x > 1

Exercise B: Answers

Reveal answers

1. x = 4
2. x = 2
3. x = 4 or x = 8
4. x = 1/5 or x = 25
5. x = 16
6. x = (1 + √33)/2

Exercise C: Answers

Reveal answers

1. x > 7
2. 1 < x < 4/3
3. x ≥ 2
4. 1 < x ≤ 9

Exercise D: Answers

Reveal answers

1. x = 1 − √17 or x = 1 + √17
2. x = 9
3. x = 4 or x = 16
4. x = 5

9. Chapter 10 Part 3: Key points to remember

1. A logarithmic argument must always be positive.
2. Combine logarithms using the product, quotient and power laws.
3. When logarithms with the same base are equal, their positive arguments are equal.
4. Use substitution when an equation is quadratic in a logarithm.
5. Use change of base when logarithms have different bases.
6. For a base greater than 1, the logarithmic function is increasing.
7. For a base between 0 and 1, the logarithmic function is decreasing.
8. Always check final answers in the original equation or inequality.

End of Part 3

You have now studied logarithmic domains, equations, inequalities and further problem solving. Continue to the next part of Chapter 10 for further applications and examination-style practice.

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